An interesting trigonometric integral

Once while solving problems from Romanian Mathematical Olympiads I encountered a mathematical gem. The problem I found was asking to calcualte the following integral I=cosxasinx+bcosxdx.I=\int\frac{\cos x}{a\sin x+b\cos x}\,dx.

What do I find especially beautiful about this problem? It rewards the desire to examine a bit more than is asked initially!

Solution. Let's try to evaluate one more integral J=sinxasinx+bcosxdx.J=\int{\frac{\sin x}{a\sin x+b\cos x}}\,dx.

Can you guess the next step?

Yes, we will take advantage of the fact that the sum of integrals is the integral of sum. In other words: aJ+bI=asinxasinx+bcosxdx+bcosxasinx+bcosxdx=asinx+bcosxasinx+bcosxdx=x+C.aJ+bI=a\int{\frac{\sin x}{a\sin x+b\cos x}}\,dx+b\int{\frac{\cos x}{a\sin x+b\cos x}}\,dx\\=\int{\frac{a\sin x+b\cos x}{a\sin x+b\cos x}}\,dx=x+C.

On other hand we can easily calculate integrals of the form f(x)f(x)dx\int{\frac{f^\prime(x)}{f(x)}}\,dx. The derivative of (asinx+bcosx) (a\sin x+b\cos x) is equal to (acosxbsinx) (a\cos x-b\sin x), which can also be expressed as a linear combination of our two integrals, i.e. aIbJ=acosxbsinxasinx+bcosxdx=lnasinx+bcosx+C.aI-bJ=\int{\frac{a\cos x-b\sin x}{a\sin x+b\cos x}}\,dx=\ln{|a\sin x+b\cos x|}+C.

Now we simply have to solve a system of linear equations. We obtain I=alnasinx+bcosx+bxa2+b2+C.I=\frac{a\ln{|a\sin x+b\cos x|}+bx}{a^2+b^2}+C. and J=axlnasinx+bcosxa2+b2+C.J=\frac{ax-\ln{|a\sin x+b\cos x|}}{a^2+b^2}+C.

Has anyone of you seen problems with the similar ideas? How often while solving a problem you investigate possible variations of the conditions?

Note by Nicolae Sapoval
5 years, 6 months ago

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Try solving this the same way: I=asinx+bcosxcsinx+dcosxdx\Large I = \int \frac {a \sin x + b \cos x }{c \sin x + d \cos x} \,dx

Pi Han Goh - 5 years, 6 months ago

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  1. Lets split the integral II into the sum of two integrals: I=asinxcsinx+dcosxdx+bcosxcsinx+dcosxdx=aJ+bK.I=a\int\frac{\sin x}{c\sin x+d\cos x}\,dx+b\int\frac{\cos x}{c\sin x+d\cos x}\,dx=aJ+bK.
  2. Now we reduced our problem to the previous one, because integrals J,KJ,K are exactly the same as the one mentioned in the post.

Nicolae Sapoval - 5 years, 6 months ago

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Pi Han Goh - 5 years, 6 months ago

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This looks very useful, thanks for sharing Nicolae. :)

Pranav Arora - 5 years, 6 months ago

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Thanks for sharing !!

Purvam Modi - 5 years, 6 months ago

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Thanks for sharing

Rajath Krishna R - 5 years, 5 months ago

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